1. Have you seen water or any other liquid poured into a narrow container? If you have, you may have noticed that the surface or water/liquid does not have a level surface but is curved. It appears concave from the top for water and most liquids but convex for mercury. Have you seen this in a thermometer or barometer? This curved surface is called the meniscus.
In chemistry, conventionally while measuring water/liquid is a burette or pipette one reads the bottom meniscus, something like this _u_, but for mercury you'd read the top -- of the meniscus. This is because these two regions are clearer and easier to read in the respective cases.
If you're wondering why this happens, the physics of surface tension can explain this in terms of the relative strengths of cohesive (same kind) and adhesive forces (opposite kind).
If the adhesive forces are stronger, the mensicus is curved upward. If the cohesive forces are stronger, the meniscus is curved downward.
2. What you're doing by closing the lid is preventing the heat loss by what a physicist would call CONVECTION. It's the loss of heat by the actual movement of the heat carried by the hot molecules.
Incidentally you're not only preventing heat loss, you're also causing an increase of pressure. This would increase the boiling point of water. In fact water boils not at 100 oC in a pressure cooker but around 120 oC due to the high pressure inside!
3. This problem involves the use of Boyle's law and pressure. Like most other problems in physics it helps to draw quick rough sketches of the set up you have.
Now the first information you have to gather is that the length of the barometer tube is 800 mm. Also the pressure of 60 mm of air space is 20 mm of mercury.
Once you have figure these two out, from your sketch you can see how
20 mm of air pressure + 740 = 760 mm of Hg
In the second case if the barometer height is only 725 mm long, then the length of the air space must be 800 - 725 = 75 mm.
Using Boyle's law you can write
20(60.A) = p(75.A)
giving you the value of p = 16 mm of Hg.
What you have done above is multiply pV on both sides, and used V = Area . height.
Writing the pressure equation like above you'll have
16 mm of air pressure + 725 = atm. pr.
that is the atmospheric pressure must now be = 741 mm of Hg.
4. I'll indicate the concepts and give you a final expression but you have to check if that's the form your teacher or the author of the textbook would be happy with.
The concepts involved here are pressure and Boyle's law.
There are three equations you can write and solve simultaneously to give the final answer in terms of Po, A, m and the acceleration due to gravity 'g'
First you must draw a diagram to imaging what's going on. Draw both the situations please next to each other.
Done that.
You know that pressure is given by the equation p = F/A = Force/Area, yeah?
Let's proceed further. From the first diagram
Po + mg/A = P1
From the second diagram
Po + xg/A = P2
where 'x' is the "weight" to be added to reduce the height to half of its initial value.
To get the THIRD equation you have to use BOYLE'S LAW. From Boyle's law, if temperature is constant
P1V1 = P2V2
Here V1 = A(h) and V2 = A(h/2)
Therefore you'll have
P1(Ah) = P2(Ah/2)
giving you as expected
P2 = 2P1
Now going back to the first two equations we have
Po + xg/A = 2(Po + mg/A)
This would give you
x = m + (PoA/g) as the weight to be added.
5.
You have to make some assumptions here. Assuming that it is a diatomic gas you're considering you can take the ratio of the specific heat capacities of the gas 'gamma' as 1.4. If it were monoatomic the value of 'gamma' would be 1.67 or triatomic would be 1.33.
We'll go with 1.4. Two equations are involved here. The first is
p(V)gamma = constant
and the second is
T(V)gamma-1 = constant
Let V, p and T be the volume, pressure and temperatures at the beginning. Let the final values be V', p' and T' respectively.
You're given that
V' = 2/3 V
or V/V' = 3/2
Subsituting in the first equation, you'll have
p'/p = (V/V')1.4
This is the ratio of the final and initial pressure values.
Before using the second equation you must convert Celsius to Kelvin.
23 0C = 300 K
Using the second equation you'll have
T(V)1.4-1 = T'(V')1.4-1
That's the final temperature.
6. This is a standard problem on gas laws involving the ideal gas equation PV = nRT. In all gas law problems the most common mistake is to forget to change temperature from Celsius to Kelvin. Please don't forget this important conversion, and other things will automatically follow.
The most important assumption here is that the two bulbs are sealed and no molecules escape during these changes. If that's the case you can reasonably assume that the total number of moles at the beginning and the end must be the same. The problem already says that the connecting tube has negligible volume.
If you make therefore the number of moles 'n' the subject of the equation, you'll have
n = PV/RT
The number of moles in the beginning would be
n = (1 atm)(600 cm3)/R.(293 K)*
I've marked the conversion to Kelvin with a * to attract your attention. The total volume would be 200 + 400 = 600 cm3 as you may have noticed. You can let pressure remain in the same units given and don't have to plug in a value for R because it would cancel in any case as you shall soon see.
The total number of moles after immersing in steam and melting ice can be seen to be
n = [(p) (400)/R (373)] + [(p) (200)/R (273)]
Equating the two equations above, cancelling and simplifying you'll find the common pressure 'p' is 0.93 atm.
7. According to Charles' law, if the mass and pressure of a gas are kept constant, the volume is directly proportional to the ABSOLUTE (or KELVIN) temperature.
That is
V proportional to T
In a hot air balloon, if the air is heated, according to Charles'law, the volume must also rise. Therefore the balloon must EXPAND. However, when the volume expands, the UPTHRUST on the balloon exerted by the surrounding air must also increase because the upthrust is PROPORTIONAL to the volume.
I think you'll like to browse through this URL's as well.
http://www.howstuffworks.com/helium.htm
It has some nice illustrations and will give you an introduction to 'How a Helium Balloon Works' and related topics at the bottom of that page.
Do you now get a better feel for 'displacement' or upthrust?
= (3/2)1.4
= 1.76
300 (V)0.4 = T'(V')0.4
T' = 300 (V/V')0.4
= 300 (3/2)0.4
= 353 K (approx to 3 sig. figs.)
= 80 0C